AFLOW Prototype: AB3_hR12_166_ac_eh-001
If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.
Links to this page
https://aflow.org/p/NHXQ
or
https://aflow.org/p/AB3_hR12_166_ac_eh-001
or
PDF Version
Prototype | BaPb$_{3}$ |
AFLOW prototype label | AB3_hR12_166_ac_eh-001 |
ICSD | 58665 |
Pearson symbol | hR12 |
Space group number | 166 |
Space group symbol | $R\overline{3}m$ |
AFLOW prototype command |
aflow --proto=AB3_hR12_166_ac_eh-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{4}, \allowbreak z_{4}$ |
GdAl$_{3}$, PuAl$_{3}$, TbAl$_{3}$, YAl$_{3}$
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Ba I |
$\mathbf{B_{2}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Ba II |
$\mathbf{B_{3}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Ba II |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Pb I |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Pb I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Pb I |
$\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |
$\mathbf{B_{8}}$ | = | $z_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |
$\mathbf{B_{9}}$ | = | $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{\sqrt{3}}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |
$\mathbf{B_{10}}$ | = | $- z_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |
$\mathbf{B_{11}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |
$\mathbf{B_{12}}$ | = | $- x_{4} \, \mathbf{a}_{1}- z_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{\sqrt{3}}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Pb II |